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Question Number 141776 by 7770 last updated on 23/May/21

$$\mathbf{The}\mathbf{volume}\mathbf{of}\mathbf{a}\mathbf{sphere}\mathbf{is}\mathbf{increasing}$$$$\mathbf{at}\mathbf{the}\mathbf{constant}\mathbf{rate}\mathbf{of}10{\mathbf{cm}}^3/\mathbf{sec}.$$$$\mathbf{Calculate}\mathbf{the}\mathbf{rate}\mathbf{of}\mathbf{increase}\mathbf{of}\mathbf{the}$$$$\mathbf{surface}\mathbf{area}\mathbf{at}\mathbf{the}\mathbf{instant}\mathbf{when}\mathbf{the}$$$$\mathbf{radius}\mathbf{is}5\mathbf{cm}.\mathbf{What}\mathbf{is}\mathbf{the}\mathbf{radius}\mathbf{of}\mathbf{the}$$$$\mathbf{sphere}\mathbf{when}\mathbf{the}\mathbf{surface}\mathbf{area}\mathbf{is}\mathbf{increasing}$$$$\mathbf{at}2{\mathbf{cm}}^2/\mathbf{sec}.$$$$\mathbf{Any}\mathbf{step}\mathbf{by}\mathbf{step}\mathbf{solution}\mathbf{pls}.$$

Answered by physicstutes last updated on 23/May/21

$$V=\frac{4}{3}\pi r^3,\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$$$$10=4\pi {\left(5\right)}^2\frac{dr}{dt}\Rightarrow \frac{dr}{dt}=\frac{10}{25\left(4\pi \right)}$$$$A=4\pi r^2\Rightarrow \frac{dA}{dt}=8\pi \frac{dr}{dt}$$

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